509. 斐波那契数

较为简单

746. 使用最小花费爬楼梯

62. 不同路径

一开始写的时候被吓到了，但是发现听完一半之后再写还是比较容易的

对于我而言主要是找到逻辑，

class Solution {
    public int uniquePaths(int m, int n) {
        if (m <= 1 || n <=1){
            return 1;
        }
        int[][] result = new int[m][n];
        result[0][0] = 0;
        result[1][0] = 1;
        result[0][1] = 1;
        for (int i = 0;i < m;i++){
            for (int j = 0;j < n; j++){
                if (i > 0 && j > 0){
                    result[i][j] = result[i][j-1] + result[i-1][j];
                }else if (i == 0 && j > 0 && j!=1){
                    result[i][j] = result[i][j-1];
                }else if (j == 0 && i > 0 && i!=1){
                    result[i][j] = result[i-1][j];
                }
        }
    }
      return result[m-1][n-1];
}
}

63. 不同路径 II

秒杀，和前者逻辑差不多，要注意初始化就好

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        if (m <= 1 || n <=1){
            for (int i = 0;i < m; i++){
                for (int j = 0;j <n; j++){
                    if (obstacleGrid[i][j] == 1){
                        return 0;
                    }
                }
            }
            return 1;
        }
        int[][] result = new int[m][n];
        if (obstacleGrid[0][0] == 1){
            return 0;
        }else{
              result[0][0] = 0;
        }

        if (obstacleGrid[1][0] == 1){
            result[1][0] = 0;
        }else{
            result[1][0] = 1;
        }
        
        if (obstacleGrid[0][1] == 1){
            result[0][1] = 0;
        }else{
            result[0][1] = 1;
        }
        
        for (int i = 0;i < m;i++){
            for (int j = 0;j < n; j++){
                if (obstacleGrid[i][j] == 1){
                    result[i][j] = 0;
                }else if (i > 0 && j > 0){
                    result[i][j] = result[i][j-1] + result[i-1][j];
                }else if (i == 0 && j > 0 && j!=1){
                    result[i][j] = result[i][j-1];
                }else if (j == 0 && i > 0 && i!=1){
                    result[i][j] = result[i-1][j];
                }
        }
    }
      return result[m-1][n-1];
}
}

别人的写法，比较nb 

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];

        //如果在起点或终点出现了障碍，直接返回0
        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {
            return 0;
        }

        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
            dp[0][j] = 1;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;
            }
        }
        return dp[m - 1][n - 1];
    }
}

 

343. 整数拆分

再写吧。。累了